Question

Find the solution of the separable equations a) the general solution: displaystyle dy/dt = t/[y^2] b) the unique solution to: dy/dt = 1/y such that if x = 3 then y = 1.

Answer 1

Answer:

(a) The general solution of given differential equation is $\frac{y^3}{3}=\frac{t^2}{2}+C$.

(b) The unique solution is $\frac{y^2}{2}=t-\frac{5}{2}$.

Step-by-step explanation:

(a)

The given differential equation is

$\frac{dy}{dt}=\frac{t}{y^2}$

Use variable separable method, to solve the above differential equation.

Separate the variables.

$y^2dy=tdt$

Integrate both sides.

$\int y^2dy=\int tdt$

$\frac{y^3}{3}=\frac{t^2}{2}+C$

The general solution of given differential equation is $\frac{y^3}{3}=\frac{t^2}{2}+C$.

(b)

The given differential equation is

$\frac{dy}{dt}=\frac{1}{y}$

Use variable separable method, to solve the above differential equation.

Separate the variables.

$ydy=1dt$

Integrate both sides.

$\int ydy=\int 1dt$

$\frac{y^2}{2}=t+C$            ... (1)

It is given that y=1 at t=3. Substitute y=1 and t=3 in the above equation.

$\frac{(1)^2}{2}=(3)+C$

$\frac{1}{2}-3=C$

$-\frac{5}{2}=C$

Substitute $C=-\frac{5}{2}$ in equation (1).

$\frac{y^2}{2}=t-\frac{5}{2}$

Therefore the unique solution is $\frac{y^2}{2}=t-\frac{5}{2}$.